Chemistry - Chemical Kinetics Question with Solution | TestHub

Order 1
Concentration changes from 0.8 M to 0.4 M in$\left(50\mathrm{\%}\right)$15 minutes this half life is = 15 minutes =${\mathrm{t}}_{50}$
A change from 0.1 M to 0.025 M is 75% and for first order reaction
${\mathrm{t}}_{75}=2\times {\mathrm{t}}_{50}=2\times 15=30$$\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}$
${\mathrm{t}}_{50}=\mathrm{15 }\mathrm{}\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}$
$\mathrm{k}=\frac{2.303\mathrm{l}\mathrm{o}\mathrm{g}2}{{\mathrm{t}}_{50}}=\frac{2.303\mathrm{l}\mathrm{o}\mathrm{g}2}{15}$
$\mathrm{\alpha }=0.1$$\mathrm{M}$
$\left(\mathrm{\alpha }-\mathrm{x}\right)=0.025$$\mathrm{M}$
for first order:
$\mathrm{k}=\frac{2.303\mathrm{l}\mathrm{o}\mathrm{g}}{\mathrm{t}}\left(\frac{\mathrm{\alpha }}{\mathrm{\alpha }-\mathrm{x}}\right)$
$\frac{2.303\mathrm{l}\mathrm{o}\mathrm{g}2}{15}=\frac{2.303}{\mathrm{t}}\mathrm{l}\mathrm{o}\mathrm{g}\frac{0.1}{0.025}=\frac{2.303}{\mathrm{t}}\mathrm{l}\mathrm{o}\mathrm{g}4$
$\frac{2.303\mathrm{l}\mathrm{o}\mathrm{g}2}{15}=\frac{2\times 2.303\mathrm{l}\mathrm{o}\mathrm{g}2}{\mathrm{t}}$
t = 30 minutes.